Answer:
The right approach will be "47° north of east".
Explanation:
The given values are:
East of prison
= 1.70 km
Displacement vector
= 2.50 km
Now,
The direction will be:
⇒ [tex]Cos \ \theta =\frac{1.7}{2.5}[/tex]
⇒ [tex]=0.68[/tex]
⇒ [tex]\theta=47.16^{\circ}[/tex]
i,e., [tex]\theta = 47^{\circ}[/tex] (north of east)
The direction of his total displacement vector with respect to due East is 1.83km
Given the following parameters
Distance ran through east = 1.70km
displacement = 2.50km
Required
Total displacement vector with respect to due north
Using the expression below;
d = √x² + y²
2.5² = 1.7² + y²
y² = 2.5² - 1.7²
y² = 3.36
y = 1.83km
Hence the direction of his total displacement vector with respect to due East is 1.83km
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